In order to prove that "[itex]\lim_{x\to a}\sqrt{x}= \sqrt{a}[/itex], you need to prove that "Given [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]|x- a|< \delta[/itex] then [itex]|\sqrt{x}-\sqrt{a}|< \epsilon[/itex]".

Start from the second inequality and show (using the above hint) that you can find such a [itex]\delta[/itex].